papaya
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complete n00b
Something is wrong with my mind:
2 NH3 + 3 O2 --> N2 + 3 H2O
The oxidant here is O2 since it goes from 0 ox. state to -2, but what is the reductant - nitrogen atom since it goes from -3 to 0 and hydrogens do
nothing? Question arises because in some EM discussions (don't remember which book) it is sometimes stated like this: "in nitro- compounds the
nitrogen is acting as a barrier between oxygens and hydrocarbon moiety which is going to be 'burned', however when you put oxidation numbers it's
obvious oxygens ''do nothing'' while the nitrogen is reduced. Another analogy - KMnO4, KClO3, etc.. "give their oxygen" in pyro compositions to burn
the fuel". Well, while these salts can decompose to give oxygen, in overall reaction with a fuel, oxygen "does nothing" (only acts as a charge carrier
in these case, if you understand what I mean).
Aren't these common misconceptions or am I a complete n00b?
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Xenoid
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Not sure what you are on about - but the equation is not balanced!
Should be;
4NH3 + 3O2 = 2N2 + 6H2O
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papaya
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Yes, you're right, didn't look carefully enough before posting, question still remains..
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Prometheus23
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With oxidizing compounds such as KMnO4 and KClO3 the atom which does the oxidizing (i.e. takes the electrons from the reducing agent) is the atom
which is bonded to the oxygens, not the oxygens themselves. For KMnO4 it is manganese which has a formal charge of +7. For KClO3 is it chlorine which
has a formal charge of +5. So you are correct that the oxygens begin with a formal charge of -2 and end up with the same charge.
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woelen
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In the case of oxidation of NH3 by O2 the hydrogens start at +1 and remain at +1. Nitrogen starts at -3 and goes to 0. Oxygen starts at 0 and goes to
-2.
if you burn a mix of e.g. KClO3 and C, you get CO2 and KCl. As many people say, the KClO3 "gives its oxygen" to burn the carbon. You could imagine it
as a two step reaction:
1) KClO3 gives KCl plus oxygen. Chlorine goes from +5 to -1, oxygen goes from -2 to 0. Potassium remains at +1.
2) Oxygen burns carbon. Carbon goes from 0 to +4, oxygen goes from 0 to -2.
So, in the end, the oxygen did not change oxidation state and charge is transferred from the reducing atom (in this case carbon) to the oxidizing atom
(in this case chlorine).
Keep in mind though that the description, given above, is a formal description. In reality the reaction most likely has another pathway from the
reactants to the products.
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